Question

In a metal wire when 10 N tensible force is applied its length becomes 5.001 m and when 20 N tensile force is applied its length became 5.002 m. So its original length is ............ .
(A) 5.001 m
(B) 4.009 m
(C) 5.0 m
(D) 4.008 m

Answer 1

Answer:

answer is a for this .......

Answer 2

Original length is 5 m

Explanation:

We know that,

$Y = \frac{\frac{10}{A}}{\frac{\Delta l}{l}}$

$Y = \frac{10\times l}{A\times\Delta l}$

$Y = \frac{20\times l}{A\times \Delta l'}$

$\frac{20\times l}{A\times\Delta l'}$ = $\frac{10\times l}{A\times \Delta l}$

$\frac{2}{\Delta l'} = \frac{1}{\Delta l}$

$2\Delta l = \Delta l'$

$l + \Delta l = 5.001 , l + \Delta l' = 5.002$

$l + 2\times \Delta l' = 5.002$

$2\times l +2\times \Delta l = 10.002$

$l = 10.002 - 5.002$

l = 5.0001 m

Original length is 5 m